= Solution
Let $M=S^2$ with an area form and let $L$ be an equator dividing the sphere into two open hemispheres of equal area. Every curve in a <symplectic surface> is Lagrangian. A small normal push moves $L$ to a nearby latitude, so it is displaceable by a <smooth isotopy>.
Suppose a <symplectic isotopy> had final image $L'$ disjoint from $L$. The curve $L'$ must lie in one hemisphere. Of the two discs bounded by $L'$, the one contained in that hemisphere has area strictly below half the total area, and the other has area strictly above half. On the other hand, a symplectomorphism maps the original two hemispheres to the two discs bounded by $L'$ and preserves their areas, so both would have half the total area. This contradiction is the <symplectic non-displaceability of an area bisector>.
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